76% people will get this right

From $76^1 = 76$ , $76^2 = 5776$ , and $76^3 = 438976$ , it would appear that the last two digits $76^n$ are $76$ for any positive integer $n$ .

This can be proved inductively:

The last two digits of the first case, $76^1 = 76$ , are $76$ .

Assuming the last two digits of $76^n$ are $76$ , then:

$76^n = 100x + 76$ (for some integer $x$ )

$76 \cdot (76^n) = 76(100x + 76)$

$76^{n + 1} = 76(100x + 76)$

$76^{n + 1} = 7600x + 5776$

$76^{n + 1} = 7600x + 5700 + 76$

$76^{n + 1} = 100(76x + 57) + 76$

$76^{n + 1} = 100x' + 76$ (for some integer $x' = 76x + 57$ )

the last two digits of $76^{n + 1}$ are also $76$ .

Therefore, the last two digits of $76^{76}$ are $\boxed{76}$ .

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Bonus: $76$ is called an automorphic number .

Let's prove that $76$ is an automorphic number as mentioned by @David Vreken, that is $76^n \equiv 76 \text{ (mod 100)}$ for natural number $n$ , using Chinese remainder theorem .

Note that $76^n \equiv 0 \text{ (mod 4)}$ and $76^n \equiv (75+1)^n \equiv 1^n \equiv 1 \text{ (mod 25)}$ . Therefore $76^n \equiv 25k + 1$ , where $k$ is an integer. Then $25k+1 \equiv 0 \text{ (mod 4)} \implies k \equiv 3$ and $76^n \equiv \boxed{76} \text{ (mod 100)}$ .

Notice that $76 \equiv -24 \pmod{100}$ . Then:

$76^{76} \equiv (-24)^{76}\pmod{100}$

But notice that $24^2 = 576$ , which is congruent to $76 \pmod{100}$ .

Extending this logic, we find that $76^{76} \equiv \boxed{76} \pmod{100}$ .

• Either by observation on putting small values of $k$ in $76^k$
• or by knowing the fact,

$76^k$ always ends with the ending two digits as $76 \quad \implies\therefore 76^{76}$ ends with $76$

The number $76$ is automorphic , hence the last two digits of $76^{76}$ would be $\boxed{{\color{#D61F06} 76}}$ .